Laplace Transform of sin(t)/t help

Applying this rule, you can differentiate under the integral sign to obtain (with the obvious re-arrangement of terms) \frac{d}{dz}\mathcal{L}\left[\frac{\sin(t)}{t}\right]=\int_0^\infty e^{-zt}\frac{\sin(t)}{t}dt. This last integral is precisely what we are interested in. We have that its value is the derivative of the Laplace transform that we wish to compute. To compute the value of the integral, note that \frac{\sin(t)}{t} can be extended by continuity at t=0 to take the value 1. Then note that in order to be able to differentiate under
  • #1
Gunthi
65
1

Homework Statement


[tex]\mathcal{L}\left[\frac{\sin(t)}{t}\right]=?[/tex]


Homework Equations


[tex]\mathcal{L}\left[f(t)\right]=\int_0^\infty e^{-zt}f(t)dt[/tex]


The Attempt at a Solution


[tex]\int_0^\infty e^{-zt}\frac{\sin(t)}{t}dt=\int_0^\infty \frac{e^{-zt}}{t}\left(\frac{e^{it}-e^{-it}}{2i}\right )dt[/tex]

[tex]\int_0^\infty f(t)dt=\lim_{T\rightarrow\infty}\int_0^T \frac{1}{t}\left(\frac{e^{t(i-z)}-e^{-t(i+z)}}{2i}\right )[/tex]

And I end up having to calculate things like
[tex]\int_0^T\frac{e^{\alpha t}}{t}dt[/tex] which doesn't seem to be the best way to do this.

Does anyone know how to do this properly?

Any help would be appreciated, thanks in advance!
 
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  • #2
First, note that as z tends to infinity, the integral approaches 0 since the kernel [itex]e^{-zt}[/itex] tends to 0 in the limit.

The relevant tool is to differentiate under the integral sign by treating the integral as a function of z. This is justified by Leibniz's rule or more advanced tools from Lebesgue integration (e.g. Lebesgue's dominated convergence theorem). In fact, one approach is to make the substitution for sin as you did, and then differentiate under the integral sign in a distributional sense, and work with Fourier transforms. But simply differentiating under the integral sign with respect to the parameter z from the beginning also suffices.
 

What is the Laplace Transform of sin(t)/t?

The Laplace Transform of sin(t)/t is 1/sqrt(s^2+1), where s is the complex variable in the Laplace domain.

Why is the Laplace Transform of sin(t)/t useful?

The Laplace Transform of sin(t)/t is useful because it allows us to solve differential equations involving sine functions, which are common in physics and engineering problems. It also simplifies the process of solving these equations by converting them into algebraic equations.

How do I calculate the Laplace Transform of sin(t)/t?

To calculate the Laplace Transform of sin(t)/t, you can use the definition of the Laplace Transform or use a table of Laplace Transform pairs. You can also use software or calculators that have the capability to perform Laplace Transforms.

What is the region of convergence for the Laplace Transform of sin(t)/t?

The region of convergence for the Laplace Transform of sin(t)/t is the entire s-plane, except for a vertical strip containing the imaginary axis. This means that the Laplace Transform of sin(t)/t exists for all values of s except for those in this vertical strip.

What are the properties of the Laplace Transform of sin(t)/t?

The Laplace Transform of sin(t)/t has the properties of linearity, time shifting, differentiation, and integration. It also has the property of convolution, which is useful in solving differential equations.

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